MECHANICS OF 2 MATERIALS - Mercer Universityfaculty.mercer.edu/kunz_rk/documents/2_1_materials_and_axial... · MECHANICS OF MATERIALS Sixth Edition Ferdinand P. Beer E. Russell Johnston,

MECHANICS OF MATERIALS

Sixth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

CHAPTER

2Stress and Strain

David F. Mazurek

Lecture Notes:

J. Walt Oler

Texas Tech University

© 2012 The McGraw-Hill Companies, Inc. All rights reserved.

– Axial Loading

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Stress & Strain: Axial Loading

• Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient.

• Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate.

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• Determination of the stress distribution within a member also requires consideration of deformations in the member.

• Chapter 2 is concerned with deformation of a structural member under axial loading. Later chapters will deal with torsional and pure bending loads.

MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf • Mazurek

Normal Strain

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strain normal

stress

==

==

L

A

P

δε

σ

Fig. 2.1

L

A

P

A

P

δε

σ

=

==22

Fig. 2.3

LL

A

P

δδε

σ

==

=

22

Fig. 2.4

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Stress-Strain Test

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Fig 2.7 This machine is used to test tensile test specimens, such as those shown in this chapter.

Fig 2.8 Test specimen with tensile load.

MECHANICS OF MATERIALS

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Stress-Strain Diagram: Ductile Materials

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MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Stress-Strain Diagram: Brittle Materials

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Fig 2.1 Stress-strain diagram for a typical brittle material.

MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf • Mazurek

Hooke’s Law: Modulus of Elasticity

• Below the yield stress

Elasticity of Modulus or Modulus Youngs=

=E

Eεσ

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• Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.

Fig 2.16 Stress-strain diagrams for iron and different grades of steel.

MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf • Mazurek

Elastic vs. Plastic Behavior

• If the strain disappears when the stress is removed, the material is said to behave elastically.

• The largest stress for which this occurs is called the elastic limit.

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• When the strain does not return to zero after the stress is removed, the material is said to behave plastically.

occurs is called the elastic limit.

Fig. 2.18

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Fatigue

• Fatigue properties are shown on S-N diagrams.

• A member may fail due to fatigueat stress levels significantly below the ultimate strength if subjected

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• When the stress is reduced below the endurance limit, fatigue failures do not occur for any number of cycles.

the ultimate strength if subjected to many loading cycles.

Fig. 2.21

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Factor of Safety

stress ultimate

safety ofFactor

u ==

=

σσ

FS

FS

Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material.

Factor of safety considerations:

• uncertainty in material properties • uncertainty of loadings• uncertainty of analyses• number of loading cycles• types of failure• maintenance requirements and

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stress allowableall==

σFS • maintenance requirements and

deterioration effects• importance of member to integrity of

whole structure• risk to life and property• influence on machine function

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Deformations Under Axial Loading

AE

P

EE === σεεσ

• From Hooke’s Law:

• From the definition of strain:

L

δε =

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• Equating and solving for the deformation,

AE

PL=δ

• With variations in loading, cross-section or material properties,

∑=i ii

iiEA

LPδFig. 2.22

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Example 2.01

psi1029 6

==

×= −E

SOLUTION:

• Divide the rod into components at the load application points.

• Apply a free-body analysis on each component to determine the internal force

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Determine the total deformation of the steel rod shown under the given loads.

in. 618.0 in. 07.1 == dD internal force

• Evaluate the total of the component deflections.

MECHANICS OF MATERIALS

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SOLUTION:

• Divide the rod into three components:

• Apply free-body analysis to each component to determine internal forces,

lb1030

lb1015

lb1060

33

32

31

×=

×−=

×=

P

P

P

• Evaluate total deflection,

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221

21

in 9.0

in. 12

==

==

AA

LL

23

3

in 3.0

in. 16

=

=

A

L

( ) ( ) ( )

in.109.75

3.0

161030

9.0

121015

9.0

121060

1029

1

1

3

333

6

3

33

2

22

1

11

−×=

×+×−+××

=

++=∑=

A

LP

A

LP

A

LP

EEA

LP

i ii

iiδ

in. 109.75 3−×=δ

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Sample Problem 2.1

The rigid bar BDE is supported by two links AB and CD.

SOLUTION:

• Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC.

• Evaluate the deformation of links ABand DC or the displacements of Band D.

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links AB and CD.

Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm2).

For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.

and D.

• Work out the geometry to find the deflection at E given the deflections at B and D.

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Sample Problem 2.1

Free body: Bar BDE

SOLUTION: Displacement of B:

( )( )( )( )

m10514

Pa1070m10500

m3.0N1060

6

926-

3

−×−=

×××−=

=AE

PLBδ

↑= mm 514.0Bδ

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( )

( )ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN300

0M

kN90

m2.0m6.0kN300

0

D

−=

×−×−=

=

+=

×+×−=

=

∑

∑↑= mm 514.0Bδ

Displacement of D:

( )( )( )( )

m10300

Pa10200m10600

m4.0N1090

6

926-

3

−×=

×××=

=AE

PLDδ

↓= mm 300.0Dδ

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Displacement of D:

( )

mm 7.73

mm 200mm 0.300mm 514.0

=

−=

=′′

x

x

x

HD

BH

DD

BB

Sample Problem 2.1

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↓= mm 928.1Eδ

( )

mm 928.1

mm 7.73

mm7.73400

mm 300.0

=

+=

=′′

E

E

HD

HE

DD

EE

δ

δ

MECHANICS OF MATERIALS

Sixth

Edition

Beer • Johnston • DeWolf • Mazurek

Saint-Venant’s Principle

• Loads transmitted through rigid plates result in uniform distribution of stress and strain.

• Stress and strain distributions

• Concentrated loads result in large stresses in the vicinity of the load application point.

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• Saint-Venant’s Principle:Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.

• Stress and strain distributions become uniform at a relatively short distance from the load application points.

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Stress Concentration: Hole

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Discontinuities of cross section may result in high localized or concentrated stresses. ave

max

σσ=K

(a) Flat bars with holes

MECHANICS OF MATERIALS

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Stress Concentration: Fillet

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(b) Flat bars with fillets

MECHANICS OF MATERIALS

Sixth

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Beer • Johnston • DeWolf • Mazurek

Example 2.12

Determine the largest axial load Pthat can be safely supported by a flat steel bar consisting of two

SOLUTION:

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b.

• Find the allowable average normal stress using the material allowable

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flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8 mm. Assume an allowable normal stress of 165 MPa.

• Apply the definition of normal stress to find the allowable load.

stress using the material allowable normal stress and the stress concentration factor.